Question #b7513

1 Answer

Answer:

#"56.74 g"#

Explanation:

Firstly, we need to determine whether the quantity of #C_2H_4# we have or the quantity of #O_2# we have is the limiting factor in how much product can be produced.

The reaction proceeds as;

#C_2H_4 + 3O_2 -> 2CO_2 + 2H_2O#

This means that for every 1 mol of ethene that reacts we need 3 mols of oxygen.

We are told that we have 32.10g of ethene. This translates to 1.15mols.

#"no. of mols" = "mass"/"Relative Atomic Mass" = "32.10g" / ("28 gmol"^-1) = "1.15 mols"#

This means that to combust 32.10g of ethene, we would need 3.45mols of Oxygen (3 x 1.15 mols). BUT we only have 64.00g of oxygen, so we need to know how many mols of oxygen we have.

#"no. of mols" = "mass"/"Relative Atomic Mass" = "64.00" / ("32 gmol"^-1) = "2 mols"#

2mols is LESS than the 3.45mols of #O_2# that we would need to completely combust 32.10g of ethene. Therefore we have to understand that not all the ethene will be reacted in this combustion - but we need to work out how many mols of ethene will react.

If we need 1 mol of ethene for 3 mols of oxygen, then 2 mols of oxygen needs 0.67mols of ethene #(2/3)#.

Now, if you look at the original reaction; 1mol of ethene reacts to produce 2 mols of #CO_2#. So in this case, 1.34mols of #CO_2# is produced (#2*0.67#).

We can now work out the mass of #CO_2# produced;

#"mass" = "number of mols" xx "relative atomic mass" = 1.34 xx 44 = "58.52 g"#

We are however told that the reaction proceeds with a #97%# yield. Therefore;

#"58.52 g" xx 0.97 = "56.74 g"#